, 2010

Suppose first that $a + b + c = 0$. Then we have
$$\begin{aligned} |a^3 b + b^3 c + c^3 a| &= |a^3 b + b^3 (-a-b) + (-a-b)^3 a| \\ &= |-b^4 - 2b^3 a - 3a^2 b^2 - 2a^3 b - a^4| \\ &= (a^2 + ab + b^2)^2. \end{aligned}$$
So it suffices to partition the set of all integers into triples of zero sum. One way to do this is the following:

a) Begin with the triple $(-1, 0, 1)$.

b) Then, let on every next step $p$ and $q$ be the least two positive integers not paired yet.

c) Form the triplets $(p, q, -p-q)$ and $(-p, -q, p+q)$ and repeat.

It is easy to see that this procedure works. $\square$